It is given that f(x) = \(\dfrac{ (4x + 3)} {(6x – 4)}\), x ≠ \( \dfrac{2}{3}\)
Now we have to show fof(x) = x
(fof)(x) = f (f(x))
=\( \dfrac{ (4x+ 3)}{(6x – 4)}\)
= \( \dfrac{34x}{34}\)
= x
Therefore, fof(x) = x for all x ≠ \( \dfrac{2}{3}\)
= fof = 1
Hence, the given function f is invertible and the inverse of f is f itself
Answered by Aaryan | 1 year agoLet A be the set of first five natural and let R be a relation on A defined as follows: (x, y) R x ≤ y
Express R and R-1 as sets of ordered pairs. Determine also
(i) the domain of R‑1
(ii) The Range of R.
A function f: R → R is defined as f(x) = x3 + 4. Is it a bijection or not? In case it is a bijection, find f−1 (3).
If f: R → R be defined by f(x) = x3 −3, then prove that f−1 exists and find a formula for f−1. Hence, find f−1 (24) and f−1 (5).
Consider f: R+ → [−5, ∞) given by f(x) = 9x2 + 6x − 5. Show that f is invertible with f-1(x) = \( \dfrac{\sqrt{(x+6)-1}}{3}\)
Consider f: R → R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is invertible with inverse f−1 of f given by f−1(x) = \( \sqrt{ (x-4)}\) where R+ is the set of all non-negative real numbers.